Class 7 Exam > Class 7 Notes > Mathematics (Maths) Class 7 > RD Sharma Solutions: Mensuration – I (Perimeter and Area of Rectilinear Figures Exercise 20.2)
Mensuration – I (Perimeter and Area of Rectilinear Figures Exercise 20.2) RD Sharma Solutions | Mathematics (Maths) Class 7 PDF Download
| Download, print and study this document offline |
Please wait while the PDF view is loading
Page 1
1. A rectangular grassy lawn measuring 40 m by 25 m is to be surrounded externally by
a path which is 2 m wide. Calculate the cost of leveling the path at the rate of Rs 8.25
per square meter.
Solution:
Let PQRS be the rectangular grassy lawn and let length = AB and breadth = BC
Given Length PQ = 40 m and breadth QR = 25 m
Area of lawn PQRS = 40 m x 25 m = 1000 m
2
Length AB = (40 + 2 + 2) m = 44 m
Breadth BC = (25 + 2 + 2) m = 29 m
Area of ABCD = 44 m x 29 m = 1276 m
2
Now, Area of the path = Area of ABCD – Area of the lawn PQRS
= 1276 m
2
– 1000 m
2
= 276 m
2
Rate of levelling the path = Rs. 8.25 per m
2
Cost of levelling the path = (8.25 x 276)
= Rs. 2277
2. One meter wide path is built inside a square park of side 30 m along its sides. The
remaining part of the park is covered by grass. If the total cost of covering by grass is
Rs 1176, find the rate per square metre at which the park is covered by the grass.
Solution:
Page 2
1. A rectangular grassy lawn measuring 40 m by 25 m is to be surrounded externally by
a path which is 2 m wide. Calculate the cost of leveling the path at the rate of Rs 8.25
per square meter.
Solution:
Let PQRS be the rectangular grassy lawn and let length = AB and breadth = BC
Given Length PQ = 40 m and breadth QR = 25 m
Area of lawn PQRS = 40 m x 25 m = 1000 m
2
Length AB = (40 + 2 + 2) m = 44 m
Breadth BC = (25 + 2 + 2) m = 29 m
Area of ABCD = 44 m x 29 m = 1276 m
2
Now, Area of the path = Area of ABCD – Area of the lawn PQRS
= 1276 m
2
– 1000 m
2
= 276 m
2
Rate of levelling the path = Rs. 8.25 per m
2
Cost of levelling the path = (8.25 x 276)
= Rs. 2277
2. One meter wide path is built inside a square park of side 30 m along its sides. The
remaining part of the park is covered by grass. If the total cost of covering by grass is
Rs 1176, find the rate per square metre at which the park is covered by the grass.
Solution:
Given that side of a square garden = 30m = a
We know that area of square = a
2
Area of the square garden including the path = a
2
= (30)
2
= 900 m
2
From the figure, it can be observed that the side of the square garden, when the path is
not included, is 28 m.
Area of the square garden not including the path = (28)
2
= 784 m
2
Total cost of covering the park with grass = Area of the park covering with green grass x
Rate per square metre
1176 = 784 x Rate per square metre
Rate per square metre at which the park is covered with grass = (1176/784)
= Rs. 1.50 per m
2
3. Through a rectangular field of sides 90 m x 60 m, two roads are constructed which
are parallel to the sides and cut each other at right angles through the center of the
field. If the width of the roads is 3 m, find the total area covered by the two roads.
Solution:
Page 3
1. A rectangular grassy lawn measuring 40 m by 25 m is to be surrounded externally by
a path which is 2 m wide. Calculate the cost of leveling the path at the rate of Rs 8.25
per square meter.
Solution:
Let PQRS be the rectangular grassy lawn and let length = AB and breadth = BC
Given Length PQ = 40 m and breadth QR = 25 m
Area of lawn PQRS = 40 m x 25 m = 1000 m
2
Length AB = (40 + 2 + 2) m = 44 m
Breadth BC = (25 + 2 + 2) m = 29 m
Area of ABCD = 44 m x 29 m = 1276 m
2
Now, Area of the path = Area of ABCD – Area of the lawn PQRS
= 1276 m
2
– 1000 m
2
= 276 m
2
Rate of levelling the path = Rs. 8.25 per m
2
Cost of levelling the path = (8.25 x 276)
= Rs. 2277
2. One meter wide path is built inside a square park of side 30 m along its sides. The
remaining part of the park is covered by grass. If the total cost of covering by grass is
Rs 1176, find the rate per square metre at which the park is covered by the grass.
Solution:
Given that side of a square garden = 30m = a
We know that area of square = a
2
Area of the square garden including the path = a
2
= (30)
2
= 900 m
2
From the figure, it can be observed that the side of the square garden, when the path is
not included, is 28 m.
Area of the square garden not including the path = (28)
2
= 784 m
2
Total cost of covering the park with grass = Area of the park covering with green grass x
Rate per square metre
1176 = 784 x Rate per square metre
Rate per square metre at which the park is covered with grass = (1176/784)
= Rs. 1.50 per m
2
3. Through a rectangular field of sides 90 m x 60 m, two roads are constructed which
are parallel to the sides and cut each other at right angles through the center of the
field. If the width of the roads is 3 m, find the total area covered by the two roads.
Solution:
Given length of rectangular field = 90m
Breadth of rectangular field = 60m
Area of the rectangular field = 90 m x 60 m = 5400 m
2
Area of the road PQRS = 90 m x 3 m
= 270 m
2
Area of the road ABCD = 60 m x 3 m
= 180 m
2
Clearly, area of KLMN is common to the two roads.
Thus, area of KLMN = 3 m x 3 m
= 9 m
2
Hence, area of the roads = Area (PQRS) + Area (ABCD) – Area (KLMN)
= (270 + 180) m
2
– 9 m
2
= 441 m
2
4. From a rectangular sheet of tin, of size 100 cm by 80 cm, are cut four squares of side
10 cm from each corner. Find the area of the remaining sheet.
Solution:
Given that length of rectangular sheet = 100 cm
Breadth of rectangular sheet = 80 cm
Area of the rectangular sheet of tin = 100 cm x 80 cm
= 8000 c m
2
Side of the square at the corner of the sheet = 10 cm
Area of one square at the corner of the sheet = (10 cm)
2
= 100 cm
2
Area of 4 squares at the corner of the sheet = 4 x 100 cm
2
= 400 cm
2
Hence, Area of the remaining sheet of tin = Area of the rectangular sheet – Area of the 4
Page 4
1. A rectangular grassy lawn measuring 40 m by 25 m is to be surrounded externally by
a path which is 2 m wide. Calculate the cost of leveling the path at the rate of Rs 8.25
per square meter.
Solution:
Let PQRS be the rectangular grassy lawn and let length = AB and breadth = BC
Given Length PQ = 40 m and breadth QR = 25 m
Area of lawn PQRS = 40 m x 25 m = 1000 m
2
Length AB = (40 + 2 + 2) m = 44 m
Breadth BC = (25 + 2 + 2) m = 29 m
Area of ABCD = 44 m x 29 m = 1276 m
2
Now, Area of the path = Area of ABCD – Area of the lawn PQRS
= 1276 m
2
– 1000 m
2
= 276 m
2
Rate of levelling the path = Rs. 8.25 per m
2
Cost of levelling the path = (8.25 x 276)
= Rs. 2277
2. One meter wide path is built inside a square park of side 30 m along its sides. The
remaining part of the park is covered by grass. If the total cost of covering by grass is
Rs 1176, find the rate per square metre at which the park is covered by the grass.
Solution:
Given that side of a square garden = 30m = a
We know that area of square = a
2
Area of the square garden including the path = a
2
= (30)
2
= 900 m
2
From the figure, it can be observed that the side of the square garden, when the path is
not included, is 28 m.
Area of the square garden not including the path = (28)
2
= 784 m
2
Total cost of covering the park with grass = Area of the park covering with green grass x
Rate per square metre
1176 = 784 x Rate per square metre
Rate per square metre at which the park is covered with grass = (1176/784)
= Rs. 1.50 per m
2
3. Through a rectangular field of sides 90 m x 60 m, two roads are constructed which
are parallel to the sides and cut each other at right angles through the center of the
field. If the width of the roads is 3 m, find the total area covered by the two roads.
Solution:
Given length of rectangular field = 90m
Breadth of rectangular field = 60m
Area of the rectangular field = 90 m x 60 m = 5400 m
2
Area of the road PQRS = 90 m x 3 m
= 270 m
2
Area of the road ABCD = 60 m x 3 m
= 180 m
2
Clearly, area of KLMN is common to the two roads.
Thus, area of KLMN = 3 m x 3 m
= 9 m
2
Hence, area of the roads = Area (PQRS) + Area (ABCD) – Area (KLMN)
= (270 + 180) m
2
– 9 m
2
= 441 m
2
4. From a rectangular sheet of tin, of size 100 cm by 80 cm, are cut four squares of side
10 cm from each corner. Find the area of the remaining sheet.
Solution:
Given that length of rectangular sheet = 100 cm
Breadth of rectangular sheet = 80 cm
Area of the rectangular sheet of tin = 100 cm x 80 cm
= 8000 c m
2
Side of the square at the corner of the sheet = 10 cm
Area of one square at the corner of the sheet = (10 cm)
2
= 100 cm
2
Area of 4 squares at the corner of the sheet = 4 x 100 cm
2
= 400 cm
2
Hence, Area of the remaining sheet of tin = Area of the rectangular sheet – Area of the 4
squares
Area of the remaining sheet of tin = (8000 – 400) cm
2
= 7600 cm
2
5. A painting 8 cm long and 5 cm wide is painted on a cardboard such that there is a
margin of 1.5 cm along each of its sides. Find the total area of the margin.
Solution:
Given length of cardboard including margin = 8 cm
Breadth of the cardboard including margin = 5 cm
Area of the cardboard including the margin = 8 cm x 5 cm = 40 c m
2
From the figure, we can observed that,
New length of the painting without margin = 8 cm – (1.5 cm + 1.5 cm)
= (8 – 3) cm
= 5 cm
New breadth of the painting without margin = 5 cm – (1.5 cm + 1.5 cm)
= (5 – 3) cm
= 2 cm
Area of the painting not including the margin = 5 cm x 2 cm = 10 cm
2
Hence, Area of the margin = Area of the cardboard including the margin – Area of the
painting
= (40 – 10) cm
2
= 30 cm
2
6. Rakesh has a rectangular field of length 80 m and breadth 60 m. In it, he wants to
make a garden 10 m long and 4 m broad at one of the corners and at another corner,
Page 5
1. A rectangular grassy lawn measuring 40 m by 25 m is to be surrounded externally by
a path which is 2 m wide. Calculate the cost of leveling the path at the rate of Rs 8.25
per square meter.
Solution:
Let PQRS be the rectangular grassy lawn and let length = AB and breadth = BC
Given Length PQ = 40 m and breadth QR = 25 m
Area of lawn PQRS = 40 m x 25 m = 1000 m
2
Length AB = (40 + 2 + 2) m = 44 m
Breadth BC = (25 + 2 + 2) m = 29 m
Area of ABCD = 44 m x 29 m = 1276 m
2
Now, Area of the path = Area of ABCD – Area of the lawn PQRS
= 1276 m
2
– 1000 m
2
= 276 m
2
Rate of levelling the path = Rs. 8.25 per m
2
Cost of levelling the path = (8.25 x 276)
= Rs. 2277
2. One meter wide path is built inside a square park of side 30 m along its sides. The
remaining part of the park is covered by grass. If the total cost of covering by grass is
Rs 1176, find the rate per square metre at which the park is covered by the grass.
Solution:
Given that side of a square garden = 30m = a
We know that area of square = a
2
Area of the square garden including the path = a
2
= (30)
2
= 900 m
2
From the figure, it can be observed that the side of the square garden, when the path is
not included, is 28 m.
Area of the square garden not including the path = (28)
2
= 784 m
2
Total cost of covering the park with grass = Area of the park covering with green grass x
Rate per square metre
1176 = 784 x Rate per square metre
Rate per square metre at which the park is covered with grass = (1176/784)
= Rs. 1.50 per m
2
3. Through a rectangular field of sides 90 m x 60 m, two roads are constructed which
are parallel to the sides and cut each other at right angles through the center of the
field. If the width of the roads is 3 m, find the total area covered by the two roads.
Solution:
Given length of rectangular field = 90m
Breadth of rectangular field = 60m
Area of the rectangular field = 90 m x 60 m = 5400 m
2
Area of the road PQRS = 90 m x 3 m
= 270 m
2
Area of the road ABCD = 60 m x 3 m
= 180 m
2
Clearly, area of KLMN is common to the two roads.
Thus, area of KLMN = 3 m x 3 m
= 9 m
2
Hence, area of the roads = Area (PQRS) + Area (ABCD) – Area (KLMN)
= (270 + 180) m
2
– 9 m
2
= 441 m
2
4. From a rectangular sheet of tin, of size 100 cm by 80 cm, are cut four squares of side
10 cm from each corner. Find the area of the remaining sheet.
Solution:
Given that length of rectangular sheet = 100 cm
Breadth of rectangular sheet = 80 cm
Area of the rectangular sheet of tin = 100 cm x 80 cm
= 8000 c m
2
Side of the square at the corner of the sheet = 10 cm
Area of one square at the corner of the sheet = (10 cm)
2
= 100 cm
2
Area of 4 squares at the corner of the sheet = 4 x 100 cm
2
= 400 cm
2
Hence, Area of the remaining sheet of tin = Area of the rectangular sheet – Area of the 4
squares
Area of the remaining sheet of tin = (8000 – 400) cm
2
= 7600 cm
2
5. A painting 8 cm long and 5 cm wide is painted on a cardboard such that there is a
margin of 1.5 cm along each of its sides. Find the total area of the margin.
Solution:
Given length of cardboard including margin = 8 cm
Breadth of the cardboard including margin = 5 cm
Area of the cardboard including the margin = 8 cm x 5 cm = 40 c m
2
From the figure, we can observed that,
New length of the painting without margin = 8 cm – (1.5 cm + 1.5 cm)
= (8 – 3) cm
= 5 cm
New breadth of the painting without margin = 5 cm – (1.5 cm + 1.5 cm)
= (5 – 3) cm
= 2 cm
Area of the painting not including the margin = 5 cm x 2 cm = 10 cm
2
Hence, Area of the margin = Area of the cardboard including the margin – Area of the
painting
= (40 – 10) cm
2
= 30 cm
2
6. Rakesh has a rectangular field of length 80 m and breadth 60 m. In it, he wants to
make a garden 10 m long and 4 m broad at one of the corners and at another corner,
he wants to grow flowers in two floor-beds each of size 4 m by 1.5 m. In the remaining
part of the field, he wants to apply manures. Find the cost of applying the manures at
the rate of Rs 300 per area.
Solution:
Given that length of a rectangular field = 80 m
Breadth of a rectangular field = 60 m
Area of rectangular field = length x breadth
= (80 x 60) m
= 4800 m
2
Again, Area of the garden = 10 m x 4 m = 40 m
2
Area of one flower bed = 4 m x 1.5 m = 6 m
2
Thus, Area of two flower beds = 2 x 6 m
2
= 12 m
2
Remaining area of the field for applying manure = Area of the rectangular field – (Area
of the garden + Area of the two flower beds)
Remaining area of the field for applying manure = 4800 m
2
– (40 + 12) m
2
= (4800 – 52) m
2
= 4748 m
2
Since 100 m
2
= 1 acre
Therefore by using the above
4748 m
2
= 47.48 acres
So, cost of applying manure at the rate of Rs. 300 per area will be = Rs. (300 x 47.48)
= Rs. 14244
7. Each side of a square flower bed is 2 m 80 cm long. It is extended by digging a strip
30 cm wide all around it. Find the area of the enlarged flower bed and also the
increase in the area of the flower bed.
Read More
|
76 videos|345 docs|39 tests
|
FAQs on Mensuration – I (Perimeter and Area of Rectilinear Figures Exercise 20.2) RD Sharma Solutions - Mathematics (Maths) Class 7
| 1. What is the formula to find the perimeter of a rectangle? |  |
Ans. The formula to find the perimeter of a rectangle is P = 2(l + b), where P is the perimeter, l is the length, and b is the breadth of the rectangle.
| 2. How can I find the area of a rectilinear figure? | |
Ans. To find the area of a rectilinear figure, divide it into smaller rectangles or squares. Calculate the area of each smaller rectangle or square using the formula A = length × breadth. Then, sum up the areas of all the smaller rectangles or squares to find the total area of the rectilinear figure.
| 3. Can you provide an example of finding the perimeter of a rectilinear figure? | |
Ans. Sure! Consider a rectilinear figure with sides measuring 5 cm, 8 cm, 3 cm, and 6 cm. To find the perimeter, add all the sides together: 5 cm + 8 cm + 3 cm + 6 cm = 22 cm. Therefore, the perimeter of the rectilinear figure is 22 cm.
| 4. How can I find the area of a rectilinear figure with curved sides? | |
Ans. If a rectilinear figure has curved sides, it is better to approximate its shape using smaller rectangles or squares. Divide the curved side into smaller straight segments and calculate the area of each segment using the formula A = length × breadth. Then, sum up the areas of all the smaller rectangles or squares to find the total approximate area of the rectilinear figure.
| 5. Is the area of a rectilinear figure always equal to its perimeter? | |
Ans. No, the area of a rectilinear figure is not always equal to its perimeter. The perimeter refers to the total length of its boundary, while the area refers to the total space enclosed by the figure. These two quantities are generally different, except for certain special cases like a square, where the area can be equal to the perimeter.
Related Exams
About this Document
|
4.65/5 Rating |
|
Nov 22, 2024 Last updated |
Document Description: RD Sharma Solutions: Mensuration – I (Perimeter and Area of Rectilinear Figures Exercise 20.2) for Class 7 2024 is part of Mathematics (Maths) Class 7 preparation.
The notes and questions for RD Sharma Solutions: Mensuration – I (Perimeter and Area of Rectilinear Figures Exercise 20.2) have been prepared according to the Class 7 exam syllabus. Information about RD Sharma Solutions: Mensuration – I (Perimeter and Area of Rectilinear Figures Exercise 20.2) covers topics
like and RD Sharma Solutions: Mensuration – I (Perimeter and Area of Rectilinear Figures Exercise 20.2) Example, for Class 7 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for RD Sharma Solutions: Mensuration – I (Perimeter and Area of Rectilinear Figures Exercise 20.2).
Introduction of RD Sharma Solutions: Mensuration – I (Perimeter and Area of Rectilinear Figures Exercise 20.2) in English is available as part of our Mathematics (Maths) Class 7
for Class 7 & RD Sharma Solutions: Mensuration – I (Perimeter and Area of Rectilinear Figures Exercise 20.2) in Hindi for Mathematics (Maths) Class 7 course.
Download more important topics related with notes, lectures and mock test series for Class 7
Exam by signing up for free. Class 7: Mensuration – I (Perimeter and Area of Rectilinear Figures Exercise 20.2) RD Sharma Solutions | Mathematics (Maths) Class 7
Description
Full syllabus notes, lecture & questions for Mensuration – I (Perimeter and Area of Rectilinear Figures Exercise 20.2) RD Sharma Solutions | Mathematics (Maths) Class 7 - Class 7 | Plus excerises question with solution to help you revise complete syllabus for Mathematics (Maths) Class 7 | Best notes, free PDF download
Information about RD Sharma Solutions: Mensuration – I (Perimeter and Area of Rectilinear Figures Exercise 20.2)
In this doc you can find the meaning of RD Sharma Solutions: Mensuration – I (Perimeter and Area of Rectilinear Figures Exercise 20.2) defined & explained in the simplest way possible. Besides explaining types of
RD Sharma Solutions: Mensuration – I (Perimeter and Area of Rectilinear Figures Exercise 20.2) theory, EduRev gives you an ample number of questions to practice RD Sharma Solutions: Mensuration – I (Perimeter and Area of Rectilinear Figures Exercise 20.2) tests, examples and also practice Class 7
tests
|
Explore Courses for Class 7 exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
Related Searches
Viva Questions
,shortcuts and tricks
,Free
,Previous Year Questions with Solutions
,Mensuration – I (Perimeter and Area of Rectilinear Figures Exercise 20.2) RD Sharma Solutions | Mathematics (Maths) Class 7
,Sample Paper
,Objective type Questions
,study material
,Semester Notes
,Extra Questions
,Exam
,mock tests for examination
,practice quizzes
,MCQs
,Important questions
,Mensuration – I (Perimeter and Area of Rectilinear Figures Exercise 20.2) RD Sharma Solutions | Mathematics (Maths) Class 7
,past year papers
,video lectures
,Summary
,Mensuration – I (Perimeter and Area of Rectilinear Figures Exercise 20.2) RD Sharma Solutions | Mathematics (Maths) Class 7
,ppt
;
Additional Information about RD Sharma Solutions: Mensuration – I (Perimeter and Area of Rectilinear Figures Exercise 20.2) for Class 7 Preparation
RD Sharma Solutions: Mensuration – I (Perimeter and Area of Rectilinear Figures Exercise 20.2) Free PDF Download
The RD Sharma Solutions: Mensuration – I (Perimeter and Area of Rectilinear Figures Exercise 20.2) is an invaluable resource that delves deep into the core of the Class 7 exam.
These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective.
With the help of these notes, you can grasp complex subjects quickly, revise important points easily,
and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner,
allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material,
or toppers' notes, this PDF has got you covered. Download the RD Sharma Solutions: Mensuration – I (Perimeter and Area of Rectilinear Figures Exercise 20.2) now and kickstart your journey towards success in the Class 7 exam.
Importance of RD Sharma Solutions: Mensuration – I (Perimeter and Area of Rectilinear Figures Exercise 20.2)
The importance of RD Sharma Solutions: Mensuration – I (Perimeter and Area of Rectilinear Figures Exercise 20.2) cannot be overstated, especially for Class 7 aspirants.
This document holds the key to success in the Class 7 exam.
It offers a detailed understanding of the concept, providing invaluable insights into the topic.
By knowing the concepts well in advance, students can plan their preparation effectively.
Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.
RD Sharma Solutions: Mensuration – I (Perimeter and Area of Rectilinear Figures Exercise 20.2) Notes
RD Sharma Solutions: Mensuration – I (Perimeter and Area of Rectilinear Figures Exercise 20.2) Notes offer in-depth insights into the specific topic to help you master it with ease.
This comprehensive document covers all aspects related to RD Sharma Solutions: Mensuration – I (Perimeter and Area of Rectilinear Figures Exercise 20.2).
It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation.
Practice papers and question papers enable you to assess your progress effectively.
Additionally, the paper analysis provides valuable tips for tackling the exam strategically.
Access to Toppers' notes gives you an edge in understanding complex concepts.
Whether you're a beginner or aiming for advanced proficiency, RD Sharma Solutions: Mensuration – I (Perimeter and Area of Rectilinear Figures Exercise 20.2) Notes on EduRev are your ultimate resource for success.
RD Sharma Solutions: Mensuration – I (Perimeter and Area of Rectilinear Figures Exercise 20.2) Class 7 Questions
The "RD Sharma Solutions: Mensuration – I (Perimeter and Area of Rectilinear Figures Exercise 20.2) Class 7 Questions" guide is a valuable resource for all aspiring students preparing for the
Class 7 exam. It focuses on providing a wide range of practice questions to help students gauge
their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation.
The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level.
Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.
Study RD Sharma Solutions: Mensuration – I (Perimeter and Area of Rectilinear Figures Exercise 20.2) on the App
Students of Class 7 can study RD Sharma Solutions: Mensuration – I (Perimeter and Area of Rectilinear Figures Exercise 20.2) alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the RD Sharma Solutions: Mensuration – I (Perimeter and Area of Rectilinear Figures Exercise 20.2),
students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of RD Sharma Solutions: Mensuration – I (Perimeter and Area of Rectilinear Figures Exercise 20.2) is prepared as per the latest Class 7 syllabus.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup